Divide. −425÷(−5) −2225 −825 825 2225

Please help me with this

attashe74 [19]

Answer:

the 1st one

Step-by-step explanation:

5 0

3 years ago

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Which best describes a transformation that preserves the size, shape, and angles of an object? A. congruent transformation B. no

Harrizon [31]

Answer:

Your answer would be D: isometry

Step-by-step explanation:

6 0

3 years ago

-x+ 2y = 10 -3x+By = 11

adoni [48]

Answer:

change the equation into one variable.

2 y equals 10 + x

y equals 5 + x by 2

now put that into the second equation

-3x + b parentheses 5 + x / 2 equals 11

-3x+b(5+ x/2)

-3x + b5 + bx/2 = 11

8 0

3 years ago

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Hellllllllllllllllllllllllllp

murzikaleks [220]

Answer:

$\begin{cases}y=-5x+1\\y=5x-4 \end{cases}$

Step-by-step explanation:

Slope-intercept form of a linear equation:

$\boxed{y=mx+b}$

where:

m is the slope.
b is the y-intercept (where the line crosses the y-axis).

Slope formula

$\boxed{\textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}}$

Equation 1

Define two points on the line:

$\textsf{Let }(x_1,y_1)=(-1, 6)$

$\textsf{Let }(x_2,y_2)=(0, 1)$

Substitute the defined points into the slope formula:

$\implies \textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{1-6}{0-(-1)}=-5$

From inspection of the graph, the line crosses the y-axis at y = 1 and so the y-intercept is 1.

Substitute the found slope and y-intercept into the slope-intercept formula to create an equation for the line:

$y=-5x+1$

Equation 2

Define two points on the line:

$\textsf{Let }(x_1,y_1)=(1, 1)$

$\textsf{Let }(x_2,y_2)=(0, -4)$

Substitute the defined points into the slope formula:

$\implies \textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-4-1}{0-1}=5$

From inspection of the graph, the line crosses the y-axis at y = -4 and so the y-intercept is -4.

Substitute the found slope and y-intercept into the slope-intercept formula to create an equation for the line:

$y=5x-4$

Conclusion

Therefore, the system of linear equations shown by the graph is:

$\begin{cases}y=-5x+1\\y=5x-4 \end{cases}$

Learn more about systems of linear equations here:

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5 0

1 year ago

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Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne

Pavlova-9 [17]

Answer:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

$p_v =P(t_{(15)}>1.507)=0.076$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X_{CTS}=2.35$ represent the mean for the sample CTS

$\bar X_{N}=1.83$ represent the mean for the sample Normal

$s_{CTS}=0.88$ represent the sample standard deviation for the sample of CTS

$s_{N}=0.54$ represent the sample standard deviation for the sample of Normal

$n_{CTS}=10$ sample size selected for the CTS

$n_{N}=7$ sample size selected for the Normal

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis: $\mu_{CTS} \leq \mu_{N}$

Alternative hypothesis: $\mu_{CTS} > \mu_{N}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{CTS}+n_{N}-2=10+7-2=15$

Since is a one side right tailed test the p value would be:

$p_v =P(t_{(15)}>1.507)=0.076$

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

6 0

3 years ago